Lemma 7.30.3. Let $\mathcal{C}$ be a site. Let $\mathcal{F}$ be a sheaf on $\mathcal{C}$. Let $\mathcal{C}/\mathcal{F}$ be the category of pairs $(U, s)$ where $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $s \in \mathcal{F}(U)$. Let a covering in $\mathcal{C}/\mathcal{F}$ be a family $\{ (U_ i, s_ i) \to (U, s)\}$ such that $\{ U_ i \to U\}$ is a covering of $\mathcal{C}$. Then $j : \mathcal{C}/\mathcal{F} \to \mathcal{C}$ is a continuous and cocontinuous functor of sites which induces a morphism of topoi $j : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/\mathcal{F}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$. In fact, there is an equivalence $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/\mathcal{F}) = \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/\mathcal{F}$ which turns $j$ into $j_\mathcal {F}$.

Proof. We omit the verification that $\mathcal{C}/\mathcal{F}$ is a site and that $j$ is continuous and cocontinuous. By Lemma 7.21.5 there exists a morphism of topoi $j$ as indicated, with $j^{-1}\mathcal{G}(U, s) = \mathcal{G}(U)$, and there is a left adjoint $j_!$ to $j^{-1}$. A morphism $\varphi : * \to j^{-1}\mathcal{G}$ on $\mathcal{C}/\mathcal{F}$ is the same thing as a rule which assigns to every pair $(U, s)$ a section $\varphi (s) \in \mathcal{G}(U)$ compatible with restriction maps. Hence this is the same thing as a morphism $\varphi : \mathcal{F} \to \mathcal{G}$ over $\mathcal{C}$. We conclude that $j_!* = \mathcal{F}$. In particular, for every $\mathcal{H} \in \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/\mathcal{F})$ there is a canonical map

$j_!\mathcal{H} \to j_!* = \mathcal{F}$

i.e., we obtain a functor $j'_! : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/\mathcal{F}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/\mathcal{F}$. An inverse to this functor is the rule which assigns to an object $\varphi : \mathcal{G} \to \mathcal{F}$ of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/\mathcal{F}$ the sheaf

$a(\mathcal{G}/\mathcal{F}) : (U, s) \longmapsto \{ t \in \mathcal{G}(U) \mid \varphi (t) = s\}$

We omit the verification that $a(\mathcal{G}/\mathcal{F})$ is a sheaf and that $a$ is inverse to $j'_!$. $\square$

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